Let
A = event that the student is on the honor roll
B = event that the student has a part-time job
C = event that the student is on the honor roll and has a part-time job
We are given
P(A) = 0.40
P(B) = 0.60
P(C) = 0.22
note: P(C) = P(A and B)
We want to find out P(A|B) which is "the probability of getting event A given that we know event B is true". This is a conditional probability
P(A|B) = [P(A and B)]/P(B)
P(A|B) = P(C)/P(B)
P(A|B) = 0.22/0.6
P(A|B) = 0.3667 which is approximate
Convert this to a percentage to get roughly 36.67% and this rounds to 37%
Final Answer: 37%
Calculate out how many times the denominator goes into the numerator. To do that, divide 234 by 8 and keep only what is to the left of the decimal point:
Since there are more parakeets than canaries, it is not possible to have only 1 of each bird in each cage <u>and</u> have the same number of birds in each cage.
He could use 42 cages, putting a canary in with the parakeet in 18 of them. Then he would have 18 cages with 2 birds each, and 24 cages with 1 bird each.
The only way to have the same number of birds (1) in all cages is to have 60 cages, 42 of which have 1 parakeet, and 18 of which have 1 canary.
_____
If more than 1 of each kind of bird can be put in the cage, the collection of birds could be put into 6 cages, each of which would be home to 7 parakeets and 3 canaries.
Answer:
All options
Step-by-step explanation:
If two coins are flipped, there are following possibilities
1. Both coins show heads
2. First coin shows heads and second shows tails
3. First coin shows tails and second shows heads
4. Both coins show tails
Let H denote heads and T denote tails
Then the sample space can be translated from the above events as:
Sample Space for Flipping Two Coins = {HH, HT, TH, TT}
which means that all the options are correct ..
Can you please add the graph, plus the question?
